# Sum of two exponential series with equal means and variances

Assuming $$A$$ and $$B$$ are two non-negative real-valued random variables such that

1. $$\mathrm{E}(A)=\mathrm{E}(B)$$ (equal means)
2. $$\mathrm{Var}(A)=\mathrm{Var}(B)<\epsilon$$ (equal small variances)

is there a way to prove that $$\frac{1}{N}\sum_{j=1}^Ne^{-a_{j}}$$ and $$\frac{1}{N}\sum_{j=1}^Ne^{-b_{j}}$$ are arbitrarily close to each other where $$a_j$$ and $$b_j$$ are realizations taken from $$A$$ and $$B$$, respectively. ($$N$$ can be assumed to be large as well)

Although it is not explicitly specified, I will assume that you intend for all the realisations of these random variables to be independent (i.e., I will assume joint independence of all the random variables in both series). The difference between the two series is the random variable defined by the function:

$$D(N) = \frac{1}{N} \sum_{i=1}^N (e^{-A_i} - e^{-B_i}).$$

Since $$e^{-a} \leqslant 1$$ for all $$a \geqslant 0$$, it follows that $$\mathbb{V}(e^{-A}) \leqslant \mathbb{E}(e^{-2A}) \leqslant 1$$ for any non-negative random variable $$A$$. Thus, we have:

\begin{aligned} \mathbb{V}(D(N)) &= \frac{1}{N^2} \sum_{i=1}^N \Big( \mathbb{V}(e^{-A_i}) + \mathbb{V}(e^{-B_i}) \Big) \\[6pt] &\leqslant \frac{1}{N^2} \sum_{i=1}^N \Big( 1 + 1 \Big) \\[6pt] &= \frac{1}{N^2} \cdot 2N \\[6pt] &= \frac{2}{N}. \\[6pt] \end{aligned}

We therefore have $$\lim_{N \rightarrow \infty} \mathbb{V}(D(N)) = 0$$, so the variance of the difference converges to zero. We also have the constant mean difference:

\begin{aligned} \mathbb{E}(D(N)) &= \frac{1}{N} \sum_{i=1}^N \Big( \mathbb{E}(e^{-A_i}) - \mathbb{E}(e^{-B_i}) \Big) \\[6pt] &= \mathbb{E}(e^{-A}) - \mathbb{E}(e^{-B}). \\[6pt] \end{aligned}

Combining these results we see that $$D(N)$$ converges in mean square to $$\mathbb{E}(e^{-A}) - \mathbb{E}(e^{-B})$$, which is a constant value. Since the variances of both random variables are small, this limiting value should be close to (but not necessarily equal to) zero. Thus, for large values of $$N$$ you would indeed expect the difference to converge towards a value near zero.

• Thank you for the answer. Is there a mathematical way to show that E(exp(-A))-E(exp(-B)) is close to zero in the end? Also, have you used anywhere the fact that means and variances of A and B are the same? – nOp May 1 '19 at 4:58
• Under the conditions you have specified, the variance of both random variables are low, so they are distributed steeply around their mean ---i.e., you have $A \approx B \approx \mu$. This gives $\mathbb{E}(e^{-A}) \approx \mathbb{E}(e^{-B})$, which means that the difference should be near zero. – Reinstate Monica May 1 '19 at 9:01
• I got back to this after a while. When you are computing the difference D(N), A_i's are realizations and therefore constant, right? Then, what does it mean when you write variance of exp(A_i), i.e., V(exp(A_i))? As variance of a constant value is zero, i can not make sense out of D(N). It would be great if you could clarify this. – nOp Jun 14 '19 at 23:22
• For the record, note that the @ben 's answer above is NOT an answer to the question. – nOp Jun 17 '19 at 18:42

Just apply the law of large numbers, since $$e^{-a_j}$$ has finite variance because $$a$$ is positive and has finite variance it self. The result is that the variance of $$e^{-a_j}$$ is smaller than the variance of $$a$$. This is elaborated in this question: https://mathoverflow.net/questions/330348/proof-of-variance-bounds-for-transformed-random-variables/330357#330357

• Thanks. LLN says the sums equal E(exp(-A)) and E(exp(-B)), respectively. What can then be said about these two means? – nOp Apr 30 '19 at 20:09
• $\frac{1}{N}\sum{exp(-a_i)}->E(exp(-A))$, when N goes to infinity – Peter Mølgaard Pallesen May 1 '19 at 8:01